Question

problem 1: a large research solenoid has a self-inductance of 2.5 h. what is the magnitude of the emf, in kilovolts, that opposes shutting off a current of 75 a through the solenoid, in 72.5 ms?

Answer 1

To calculate the induced emf in the solenoid, we use the formula ε = -L(ΔI/Δt). The magnitude of the emf that opposes shutting off a current of 75 A through the solenoid with a self-inductance of 2.5 H in 72.5 ms is 2.5862 kilovolts.

Step-by-step explanation:

The student is asking about the magnitude of the electromotive force (emf) induced in a solenoid when the current through it is switched off. This is a problem that involves the concept of self-inductance in physics. The formula we use to calculate the induced emf (ε) when the current through inductor changes is ε = -L(ΔI/Δt), where L is the self-inductance, ΔI is the change in current, and Δt is the time over which this change occurs.

Using the given values: L = 2.5 H, ΔI = -75 A (since the current is turning off, we use the negative sign to indicate a decrease), and Δt = 72.5 ms = 0.0725 s, we find that the magnitude of the induced emf is:

ε = |-L(ΔI/Δt)| = |-(2.5 H)(-75 A / 0.0725 s)|

ε = |(2.5 H)(75 A / 0.0725 s)|

ε = |(2.5 H)(1034.48 A/s)|

ε = |2586.2 V| or 2.5862 kV

Therefore, the magnitude of the emf that opposes shutting off the current is 2.5862 kilovolts.